### All in the Mind

Imagine you are suspending a cube from one vertex (corner) and allowing it to hang freely. Now imagine you are lowering it into water until it is exactly half submerged. What shape does the surface of the water make around the cube?

### Just Rolling Round

P is a point on the circumference of a circle radius r which rolls, without slipping, inside a circle of radius 2r. What is the locus of P?

### Coke Machine

The coke machine in college takes 50 pence pieces. It also takes a certain foreign coin of traditional design...

# Hamiltonian Cube

##### Stage: 3 and 4 Short Challenge Level:

The image above shows a possible path. Each edge joining a corner to a face centre has length $\frac{1}{\sqrt{2}}$ (by Pythagoras' Theorem), while each edge which joins two adjacent corners has length $1$. So the length of the path above is $1+6\sqrt{2}$. This is the length of the shortest path to pass through all the vertices.

To prove this, note the length of the shortest path must be at least $\frac{13}{\sqrt{2}}$. Such a path would move alternately between corners and faces, but as there are $8$ corners and only $6$ faces, so this is impossible. So at least one of the edges must join to corners, and so the shortest length is $1+6\sqrt{2}$.

This problem is taken from the UKMT Mathematical Challenges.